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NUCLEAR STRUCTURE OF Xe-124, Xe-126, Xe-128, Xe-129, Xe-130, Xe-131, Xe-132, Xe-134
By Prof. Lefteris Kaliambos (Λευτέρης Καλιαμπός) Τ.Ε. Institute of Larissa Greece ( July 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity led to the abandonment of electromagnetic laws in favor of wrong theories which could not lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give the nuclear binding and nuclear structure by applying the well-established laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). Nuclear structure of Xenon with 30 blank positions Naturally occurring xenon (Xe) is made of 8 stable isotopes. (Xe-124, Xe-126, and Xe-134 are predicted to undergo double beta decay, but this has never been observed in these isotopes, so they are considered to be stable. Comparing the Xenon of 54 protons with Tellurium of 52 protons (even number ) we conclude that the structure of Xenon has the same high symmetry as that of Tellurium. ( See my STRUCTURE OF Te-120..Te-130 ). In the following diagram of Xe the additional p53n53 and p54n54 are not shown. However according to the top view of the third plane of Xe the p53 and n54 replace the 1(p) formed by n42 and n47 and the 1(n) formed by p44 and p48. (Top view of the third h. plane of Te). Note that these two symmetrical pn vertical systems fill the blank positions of the third and the fourth plane in order to make two symmetrical vertical rectangles Under this condition the number N of blank positions is given by The two squares give 8n The first and sixth plane give 4(n) The second and fifth plane give 4{n} +8n The third and the fourth plane give the rest 6(n) That is N = 8n + 4(n) + 4{n} + 8n + 6(n) = 30 extra neutrons of opposite spins Or N = 4{n} + 16n + 10(n) = 30 extra neutrons of opposite spins ' ' STRUCTURE OF Xe-124, Xe-126, Xe-128, Xe-130, Xe-132 Xe-134 WITH S = 0 Since the 54 protons and 54 neutrons give S=0 one concludes that the S=0 of the above stable nuclide is due to the number of extra neutrons with opposite spins . For example the Xe-124 with 16 extra neutrons has 4{n} + 12n with opposite spins while the XE-134 with 26 extra neutrons has 4{n} + 16n + 6(n) with opposite spins. STRUCTURE OF Xe-129 WITH S=+1/2 AND OF Xe-131WITH S = +3/2 Since the total spin of Xe-129 is due to the total spin +1/2 of the extra neutrons we conclude that the Xe- 129 of 21 extra neutrons with the spin S=+1/2 has 11 extra neutrons of positive spins and 10 extra neutrons of negative spin. Whereas the Xe-131 of 23 extra neutrons with the total spin S = +3/2 has 13 extra neutrons of positive spins and 10 extra neutrons of negative spins. DIAGRAM OF Xe FORMING 30 BLANK POSITIONS Here the additional p53n53 and p54n54 ara not shown. You can see the p47n47 along with the p48n48, which make two symmetrical alpha particles of opposite spins . But you cannot see the additional p49n49 the n39p39 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. Also the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Moreover the extra neutrons 4(n) of the first and the sixth plane are not shown, while the extra neutrons 4n existing over the p31 and p32, and under the p21 and p22 with the extra 4n existing near the p23, p24, p29 and p30, are shown. ' n40.......p40.......n ' ' n........p38..........n38 H. Square with n' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 Sixth H. plane' ' n........ p29.........n10.........p10…… n30' ' n29………p9..........n9 …….p30..........n Fifth H. plane' ' p47.......n27.........p8............n8.........p28........n48' ' n45.......p27.........n7..........p7.........n28..........p46 Fourth H. plane' ' n47......p25.........n6.........p6..........n26...........p48' ' p45......n25……….p5..........n5……….p26.........n46 Third H. plane' ' n23……p4..........n4………….p24...........n' ' n........p23……....n3………p3………..n24 Second H.plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First H. plane' ' n.......p37......n37 ' ' n39.....p39..........n ' H. Square with n TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' (n).......p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' ' ' TOP VIEW OF THE SECOND HORIZONTAL PLANE Here the n near the p14 fills the blank position formed by p51 and p14. While the {n} near the p14 fills the blank position formed by p14, p24 and p44. The 2n near p24 and p23 fill the blank positions formed by p24 and p48 as well as by p23 and p45. Moreover the blank position of {n} near p23 is formed by p23, p13 and p41. Finally the blank position of n near the p13 is formed byp13 and p49. That is we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. ' n' ' n14.......p14........{n}' ' n23.......p4........n4.........p24......n' n.......p23........n3........p3.........n24 ' {n}.......p13......n13' ' n ' ' ' ' ' ' ' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' '''HERE YOU SEE THE p53 AND n54 WHICH REPLACE THE 1(p) AND 1(n) OF Te. SO THE NUMBER (n) OF THE THIRD AND THE FOURTH PLANE IS 6(n). HERE THE p41, n42, n43 AND p44 MAKE THE SQUARE OF THE CENTRAL PARALLELEPIPED, WHILE THE p45 AND n47 ALONG WITH THE n46 AND p48 ARE THE DEUTERONS OF THE TWO ALPHA PARTICLES. AT THE SAME PLANE ARE SHOWN ALSO THE DEUTERONS LIKE n16p16 AND p15n15 ALONG WITH THE ADDITIONAL p49, n52, n50, AND p51 WHICH ARE THE DEUTERONS OF THE TWO ALPHA PARTICLES EXISTING IN FRONT OF THE CENTRAL PARALLELEPIPED AND BEHIND IT. NOTE THAT THE REST 3(n) AND 3(p) WILL BE USED FOR THE CONSTRUCTION OF THE NEXT HEAVIER NUCLEI. ' (p)........n50.......p51......(n) ' ' p53........n42........p16......n16......p44.........n54''' ' n47........p25........n6........p6........n26.........p48' ' p45........n25........p5........n5........p26........ n46' ' (n).........p41.......n15.......p15.......n43.........(p)' ' (n)........p49.......n52.......(p)' ' ' Category:Fundamental physics concepts